# Consider the reaction Cr2O7^2− +14H^+ + 6e^− →2Cr^3+ + 7H2O. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O7^2-? (A) 6 x 10^6 C (B) 5.79 x 10^5 C (C) 5.25 x 10^5 C (D) None of these

Given reaction is

$$Cr_{2}O_{7}^{2-}+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+7H_{2}O$$

1 F = 96500 C = 1 mole of e

1 mole of $$Cr_{2}O_{7}^{2-}$$ → 6 moles of e

1 mole of e = 96500 C

6 moles of e = 6 x 96500 C

We get,

6 moles of e = 579000 C

6 moles of e = 5.79 x 105 C

Therefore, the required quantity of electricity is 5.79 x 105 C

So, the correct option is (B)